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Ellipse question

2022 · 27 Jun · Shift 1 · Q30

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Ellipse
2022 · 27 Jun · Shift 1 · Q30

Ellipse question

2022 · 27 Jun · Shift 1 · Q30

JEE MainMathematicsEllipseMCQ+4 / −1

Let the eccentricity of an ellipse

{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

,

a \gt b

, be

{1 \over 4}

. If this ellipse passes through the point

\left( { - 4\sqrt {{2 \over 5}} ,3} \right)

, then

{a^2} + {b^2}

is equal to :

A
29
B
31
C
32
D
34

View written solutionFree

Correct answer: B

Use the eccentricity condition

For the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b,

the eccentricity is

e=\sqrt{1-\frac{b^2}{a^2}}.

Given

e=\frac14,

so

1-\frac{b^2}{a^2}=\frac1{16}.

Hence,

\frac{b^2}{a^2}=1-\frac1{16}=\frac{15}{16}.

Thus,

b^2=\frac{15}{16}a^2.

Use the fact that the ellipse passes through the given point

The point is

\left(-4\sqrt{\frac25},\,3\right).

So

x^2=\left(-4\sqrt{\frac25}\right)^2=16\cdot\frac25=\frac{32}{5},

and

y^2=3^2=9.

Substitute into the ellipse equation:

\frac{32/5}{a^2}+\frac{9}{b^2}=1.

Using

b^2=\frac{15}{16}a^2,

we get

\frac{32}{5a^2}+\frac{9}{(15/16)a^2}=1.

Now simplify the second term:

\frac{9}{(15/16)a^2}=\frac{9\cdot 16}{15a^2}=\frac{144}{15a^2}=\frac{48}{5a^2}.

Therefore,

\frac{32}{5a^2}+\frac{48}{5a^2}=1.

So,

\frac{80}{5a^2}=1 \quad\Rightarrow\quad \frac{16}{a^2}=1.

Hence,

a^2=16.

Then

b^2=\frac{15}{16}\cdot 16=15.

Find $a^2+b^2$

a^2+b^2=16+15=31.

Check options

The correct option is

\boxed{31}

which is Option B.

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